Question

The number of terms of an A.P. is even ; sum of all terms at odd places and even places are $24$ and $30$ respectively . Last term exceeds the first term by $10.5$. Then the number of terms in the series is

• A. $8$
• B. $12$
• C. $16$
• D. $20$

Answer 1

The correct option is B $12$

$\begin{array}{c}\text{Given that}\\ \begin{array}{cc}\pi ,& T_2,T_3\dots ..T_{2n}\text{are in}AP\\ T_1+T_3+T_5+\dots +T_{2n-1}=24& \\ T_2+T_4+T_6+\dots +T_{2n}=30& \end{array}\\ T_{2n}=T_1+10.5\\ \text{let}{T}_1=a\text{and}d=\text{common difference.}\end{array}$

$\begin{array}{cc}\Rightarrow & a+(2n-1)d=a+10.5\Rightarrow (2n-1)d=10.5\\ T_1+T_3+T_5+\dots +T_{n-1}& =24\end{array}$

$\begin{array}{c}\Rightarrow a+(a+2d)+(a+4d)+\cdots +(a+(2n-2\left)d\right)=24\\ na+[0+2d+4d+\dots +(2n-2\left)d\right]=24\\ na+2d[1+2+3+\dots +(n-1\left)\right]=24\end{array}$

$\begin{array}{c}na+2d\left[\frac{(m-1)(n+1)}{2}\right]=24\\ \Rightarrow na+d(n-1)(n+0)=24\Rightarrow n(a+n-1)d=24-\left(1\right)\end{array}$

$\begin{array}{c}{T}_2+T_4+T_6+\dots +T_{2n}=30\\ \Rightarrow (a+d)+(a+3d)+(a+5d)+\dots +(a+(2n-1\left)d\right)=30\\ na+d(1+3+5+\dots +2n-1)=30\\ na+d{n}^2=30\end{array}$

$\begin{array}{c}n(a+dn)=30\\ \text{Fuom}\left(1\right)\text{and}\left(2\right)\\ na+d{n}^2-dn=24\\ 30-d{n}^2=24\\ d{n}^2=6\Rightarrow n^2=\frac{6(2n-1)}{10.5}\\ \Rightarrow n=12\end{array}$

Hence B is the correct option